Class 9: Perimeter and Areas of Plane Figures – Exercise 15.4 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the perimeter and area of a rectangle whose length and breadth are $20 \ cm$ and $8 \ cm$ respectively.

Answer:

Dimensions of the rectangle: Length $(l) = 20 \ cm$ Breadth $(b) = 8 \ cm$

Therefore Perimeter $= 2 (l + b) = 2 ( 20 + 8) = 56 \ cm$

Area $= l \times b = 20 \times 8 = 160 \ cm^2$

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Question 2: A rectangular room floor is $192 \ m^2$ in area. If its length is $16 \ m$ , find its perimeter.

Answer:

Dimensions of the rectangle: Length $(l) = 16 \ m$ Let Breadth $(b) = x \ m$

Area of rectangle $= l \times b$

$\Rightarrow 192 = 16 \times x \Rightarrow x = 12$

Hence Breadth $= 12 \ m$

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Question 3: Find the length of a diagonal of a rectangle whose adjacent sides are $8 \ m$ and $6 \ m$ long.

Answer:

Dimensions of the rectangle: Length $(l) = 8 \ m$ Let Breadth $(b) = 6 \ m$

Diagonal of a rectangle $= \sqrt{l^2 + b^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10 \ m$

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Question 4: Find the length of a diagonal of a square of side $4 \ cm$ .

Answer:

Dimension of a square: Side $(a) = 4 \ cm$

Diagonal of a square $= \sqrt{2} a = 4 \sqrt{2} \ cm$

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Question 5: Find the perimeter of a square the sum of the lengths of whose diagonal is $144 \ cm$ .

Answer:

Dimension of a square: Side $= a$

Given: Diagonal of the square $= 144 \ cm$

We know diagonal of a square $= \sqrt{2} a$

$\Rightarrow 2 (\sqrt{2} a) = 144$

$\Rightarrow a =$ $\frac{72}{\sqrt{2}}$ $= 36\sqrt{2} \ cm$

Perimeter of a square $= 4a = 4 \times 36\sqrt{2} = 144\sqrt{2} \ cm$

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Question 6: The length and breadth of a room are in the ratio $3 : 2$ . Its area is $216 \ m^2$ . Find its perimeter.

Answer:

Dimensions of the rectangle: Let Length $(l) = 3x$ Let Breadth $(b) = 2x$

Given: Area is $216 \ m^2$

Therefore $216 = 3x \times 2x$

$\Rightarrow x^2 = 36$

$\Rightarrow x = 6$

Hence Length $(l) = 18 \ m$ Let Breadth $(b) = 12 \ m$

Therefore Perimeter $= 2 (l + b) = 2 (18 + 12) = 60 \ cm$

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Question 7: The diagonal of a square $A$ is $(a + b)$ . Find the diagonal of a square $B$ whose area is twice the area of $A$ .

Answer:

Given: Diagonal of a square $A$ is $(a + b)$

If the side of the square $= x$

$\Rightarrow Diagonal = \sqrt{2} x = (a+b)$

$\Rightarrow x =$ $\frac{a+b}{\sqrt{2}}$ … … … … … (i)

Let the side of the second square $= y$

Given: $y^2 = 2 x^2 \Rightarrow y = \sqrt{2} x$

Diagonal $= \sqrt{2} y = \sqrt{2} \times \sqrt{2} x = 2x$ … … … … … (ii)

Substituting (i) in (ii) we get

Diagonal $= 2 \times \Big($ $\frac{a+b}{\sqrt{2}}$ $\Big) = \sqrt{2} (a+b)$

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Question 8: The perimeter of a square is $(4x + 20) \ cm$ . Find its diagonal.

Answer:

Perimeter $= 4x + 20$

$\Rightarrow Side =$ $\frac{4x+20}{4}$ $= x+5$

Hence Diagonal $= \sqrt{2} (x+5)$

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Question 9: Find the area of a square that can be inscribed in a circle of radius $10 \ cm$ .

Answer:

Radius $= 10 \ cm$

Hence the length of the side $= \sqrt{10^2 + 10^2} = 10 \sqrt{2}$

Therefore Area $= 10 \sqrt{2} \times 10 \sqrt{2} = 200 \ cm^2$

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Question 10: Find the perimeter of a square the sum of the lengths of whose diagonals is $100 \ cm$ .

Answer:

Let the side be $= x$

Given: $\sqrt{2} x + \sqrt{2} x = 100$

$\Rightarrow 2 \sqrt{2} x = 100$

$\Rightarrow x =$ $\frac{50}{\sqrt{2}}$ $= 25 \sqrt{2}$

Therefore Perimeter $= 4 \times 25 \sqrt{2} = 100 \sqrt{2} \ cm$

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Question 11: The diagonal of a square is $14 \ cm$ . Find its area.

Answer:

Let the side of the square $= x$

Given: $\sqrt{2} x = 14$

$\Rightarrow x = 7 \sqrt{2}$

Therefore Area $= 7 \sqrt{2} \times 7 \sqrt{2} = 28 \ cm^2$

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Question 12: Find the area and perimeter of a square plot of land the length of whose diagonal is $15 \ m$ .

Answer:

Given: Diagonal $= 15 \ m$

Let the side of the square $= x$

Therefore $\sqrt{2} x = 15$

$\Rightarrow x =$ $\frac{15}{\sqrt{2}}$

Therefore Area $=$ $\frac{15}{\sqrt{2}}$ $\times$ $\frac{15}{\sqrt{2}}$ $=$ $\frac{225}{2}$ $= 112.5 \ m^2$

Hence the Perimeter $= 5 \times$ $\frac{15}{\sqrt{2}}$ $= 30 \sqrt{2}$

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Question 13: Find the ratio of the area of a square to that of the square drawn on its diagonal.

Answer:

Let the side of square $= x$

Therefore diagonal $= \sqrt{2} x$

Hence the ratio $=$ $\frac{x^2}{(\sqrt{2} x)^2}$ $=$ $\frac{1}{2}$

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Question 14: The diagonal of square $A$ is $(a + b)$ . Find the diagonal of square $B$ whose area is half of the area of $A$ .

Answer:

Let the side of square $A = x$

Therefore $\sqrt{2} x = a+b$

$\Rightarrow x =$ $\frac{a+b}{\sqrt{2}}$

Let the side of square $B = y$

Therefore $y^2 =$ $\frac{1}{2}$ $x^2$

$\Rightarrow y =$ $\frac{1}{\sqrt{2}}$ $x$

Diagonal of square $B = \sqrt{2} y = \sqrt{2} \times$ $\frac{1}{\sqrt{2}}$ $x =$ $\frac{a+b}{\sqrt{2}}$

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Question 15: The perimeter of a square is $48 \ m$ . The area of a rectangle is $4 \ sq. m$ less than the area of the given square. If the length of the rectangle is $14 \ m$ , find its breadth.

Answer:

Let the side of the square $= a$

Therefore $4a = 48 \Rightarrow a = 12 \ m$

Let the breadth of the rectangle $= b$

Therefore $14 \times b = 12^2 - 4$

$\Rightarrow 14b = 144 - 4$

$\Rightarrow b = 10 \ m$

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Question 16: The perimeter of one square is $748 \ cm$ and that of another is $336 \ cm$ . Find the perimeter and the diagonal of a square whose area is equal to the sum of the areas of these two squares.

Answer:

Let the side of square 1 $= a_1$

Therefore 4 $a_1 = 748 \Rightarrow a_1 = 187 \ cm$

Hence the area of square 1 $= 187^2 = 34969 \ cm^2$

Let the side of square 2 $= a_2$

Therefore 4 $a_2 = 336 \Rightarrow a_2 = 84 \ cm$

Hence area of square 2 $= 84^2 = 7056 \ cm^2$

Therefore area of square 3 $= 34969 + 7056 = 42025 \ cm^2$

Hence the side of square 3 $= \sqrt{42025} = 205 \ cm$

Therefore the perimeter of square 3 $= 4 \times 205 = 820 \ cm$

Diagonal of square 3 $= \sqrt{2} (205) = 289.91 \ cm$

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Question 17: The perimeter of a rectangular card board is $96 \ cm$ . If its breadth is $18 \ cm$ , find the length and area of the card board.

Answer:

Dimensions of the rectangle: Length $(l) = l$ Breadth $(b) = 18 \ cm$

Therefore $2(l+b) = 96$

$\Rightarrow l+18 = 48$

$\Rightarrow l = 30 \ cm$

Hence area $= 30 \times 18 = 540 \ cm^2$

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Question 18: If the sides of two squares are in the ratio $x : y$ , prove that their areas are in the ratio $x^2:y^2$ .

Answer:

Side of square 1 $= x$

Therefore Area of square 1 $= x^2$

Side of square 2 $= y$

Therefore Area of square 2 $= y^2$

Therefore ratio of areas $=$ $\frac{x^2}{y^2}$

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Question 19: In exchange for a square plot one of whose sides is $84 \ m$ , a man wants to buy a rectangular plot $144 \ m$ long and of the same area as of the square plot. Find the width of the rectangular plot.

Answer:

Side of square plot $= 84 \ m$

Length of rectangular plot $= 144 \ m$

Let breadth of rectangular plot $= b$

Therefore $144 \times b= 84 \times 84$

$\Rightarrow b = 49 \ m$

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Question 20: A rectangular lawn $80 \ m \times 60 \ m$ has two roads each with $10 \ m$ wide running in the middle of it, one parallel to the length and other parallel to the breadth. Find the cost of graveling them at $30$ paisa per square meter.

Answer:

Area of graveled road $= 10 \times 80 + 10 \times 60 - 10 \times 10 = 800 + 600 - 100 = 1300 \ m^2$

Therefore cost of graveling $= 1300 \times 0.30 = 390 \ Rs.$

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Question 21: The area of a square plot is $\frac{1}{2}$ hectare. Find the diagonal of the square.

Answer:

Let the side of square plot $= a$

Therefore $a^2 = \frac{1}{2}$ hectare

We know 1 hectare $= 10000 \ m^2$

Therefore $a^2 = 5000 \Rightarrow a = 50 \sqrt{2}$

Hence diagonal $= \sqrt{2} a = \sqrt{2} (50 \sqrt{2}) = 100 \ m$

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Question 22: A lawn is in the form of a rectangle having its sides in the ratio $5: 2$ . The area of the lawn is $1000 \ m^2$ . Find the cost of fencing it at the rate of $Rs. \ 8.50$ per meter.

Answer:

Let length $= 5x$ and breadth $= 2x$

Therefore $5x \times 2x = 1000 \Rightarrow 10x^2 = 1000 \Rightarrow x = 10 \ m$

Therefore length $= 50 \ m$ and breadth $= 20 \ m$

Therefore perimeter $= 2 (50 + 20) = 140 \ m$

Therefore cost of fencing $= 140 \times 8.5 = 1190 \ Rs.$

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Question 23: The area of a square park is $40,000 \ sq. m$ . Find the cost of fencing it at the rate of $Rs. \ 2.80$ per meter.

Answer:

Area of square $= 40000 \ m^2$

Let side of the square $= a$

$\Rightarrow a^2 = 40000 \Rightarrow a = 200 \ m$

Therefore Perimeter $= 4a = 800 \ m$

Therefore cost of fencing $= 800 \times 2.8 = 2240 \ Rs.$

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Question 24: The area of the base of a rectangular tank is $2400 \ m^2$ and its sides are in the ratio $3: 2$ . Find the cost of planting flowers round it at the rate of $Rs. \ 1.25$ per meter.

Answer:

Let length $= 3x$ and breadth $= 2x$

Therefore $3x \times 2x = 2400 \Rightarrow x^2 = 400 \Rightarrow x = 20 \ m$

Therefore length $= 60 \ m$ and breadth $= 40 \ m$

Therefore perimeter $=2 (60 + 40) = 200 \ m$

Therefore cost of planting flowers $= 200 \times 1.25 = 250 \ Rs.$

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Question 25: A rectangular field $242$ meter long has got an area of $4840 \ sq. m$ , what will be the cost of fencing that field on all the four sides, if $1$ meter of fencing costs $20$ paisa?

Answer:

Length $= 242 \ m$

Area $= 4840 \ m^2$

Therefore breadth $=$ $\frac{4840}{242}$ $= 20 \ m$

Perimeter $= 2 (242 + 20) = 524 \ m$

Therefore cost of fencing $= 524 \times 0.2 = 1048 \ Rs.$

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Question 26: A rectangular grassy plot is $112 \ m \times 78 \ m$ . It has gravel path $2.5 \ m$ wide all around it on the inside. Find the area of the path and the cost of constructing it at the rate of $Rs. \ 3.40$ per sq. meter.

Answer:

Dimensions of park: Length $= 112 \ m$ , Breadth $= 78 \ m$

Inner Dimensions of park: Length $= (112 - 5) = 107 \ m$ , Breadth $= (78-5) = 73 \ m$

Area of path $= 112 \times 78 - 107 \times 73 = 925 \ m^2$

Therefore cost of construction $= 925 \times 3.4 = 3145 \ Rs.$

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Question 27: There is a square field whose side is $44 \ m$ . A flowerbed is prepared in its center, leaving a gravel path of uniform width all around the flower bed. The total cost of laying the flower bed and graveling the path at $Rs. \ 2$ and $Rs. \ 1$ per square meter respectively is $Rs. \ 3536$ . Find the width of the gravel path.

Answer:

Let the side of the garden $= x$

Therefore the area of the garden $= x^2$

Area of the path $= 44^2 - x^2$

Therefore $3536 = 2 \times x^2 + 1 \times ( 44^2 - x^2)$

$\Rightarrow 3536 = 2x^2 + 44^2 - x^2$

$\Rightarrow x^2 = 3536 - 44^2 = 1600$

$\Rightarrow x = 40 \ m$

Hence the path $= 2 \ m \ wide$

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Question 28: How many tiles $40 \times 40 \ cm^2$ each will be required to pave the footpath $1 \ m$ wide carried round the outside of a grassy plot $28 \ m$ by $18 \ m$ ?

Answer:

Are of footpath $= 20 \times 30 - 28 \times 18 = 96 \ m^2$

No of times required $=$ $\frac{96}{0.4 \times 0.4}$ $= 600 \ tiles$

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Question 29: A room is $7.4 \ m$ long, $3.6 \ m$ broad and $4 \ m$ high. It has two doors $2.1 \ m \times 1.2 \ m$ and $5$ windows each $1.8 \ m \times 1.2 \ m$ . How much will it cost to whitewash the walls of the room at the rate of $Rs.\ 2.00$ per square meter?

Answer:

Dimensions of the Room: Length (l) $= 7.4 \ m$ , Breadth $(b) = 3.6 \ m$ and Height $(h) = 4 \ m$

Dimensions of the Doors: Breadth $(b) = 2.1 \ m$ and Height $(h) = 1.2 \ m$

Dimensions of the Window: Breadth $(b) = 1.8 \ m$ and Height $(h) = 1.2 \ m$

Area of walls = Lateral Area $= 2 (l+b) h = 2 (7.4 + 3.6) \times 4 = 88 \ m^2$

Area of Doors and Windows $= 2 \times 2.1 \times 1.2 + 5 \times 1.8 \times 1.2 = 15.84 \ m^2$

Area to be painted $= 72.16 \ m^2$

Cost of painting $= 72.16 \times 2 = 144.32 \ Rs.$

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Question 30: A carpet is laid on the floor of a room $8 \ m \times 5 \ m$ . There is a border of constant width around the carpet. If the area of the border is $1.2 \ m^2$ , find its width.

Answer:

Dimensions of the Room: Length (l) $= 8 \ m$ , Breadth $(b) = 5 \ m$

Area of Border $= 14 \ m^2$

Therefore $14 = (5+2x)(8+2x) - 8 \times 5$

$\Rightarrow 12 = 40 + 16x + 10x +4x^2 - 40$

$\Rightarrow 4x^2 + 26x^ - 14 = 0$

$\Rightarrow 2x^2 + 13x - 7 = 0$

$\Rightarrow (2x-1)(x+7) = 0$

$\Rightarrow x =$ $\frac{1}{2}$ or $x = -7$ (this is not possible as $x$ cannot be negative)

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Question 31: A rectangular courtyard, $3.78 \ m$ long and $5.25 \ m$ broad, is to be paved exactly -with square tiles, all of the same size. Find the largest size of such a tile and the number of tiles required to pave it.

Answer:

Dimension of courtyard: Length $(l) = 3.78 \ m \ or \ 378 \ cm$

Breadth $(b) = 5.25 \ m \ or \ 525 \ cm$

The largest common divisor of both the numbers is $21$

Therefore the dimension of the largest square tile is $21 \ cm$

Hence the number of tiles required $=$ $\frac{3.78 \times 5.25}{0.21 \times 0.21}$ $= 450$

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Question 32: The cost of fencing a square field at $60$ paisa per meter is $Rs. \ 1200$ . Find the cost of reaping the field, it the rate of $50$ \ paisa per $100 \ sq. m$ .

Answer:

Perimeter $\times 0.60 = 1200$

$\Rightarrow Perimeter = 2000 \ m$

Therefore side $=$ $\frac{2000}{4}$ $= 500 \ m$

Therefore Area $= 500 \times 500 = 250000$

Therefore cost of reaping the field $=$ $\frac{0.50}{100}$ $\times 250000 = 1250 \ Rs.$

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Question 33: A room $4.9 \ m$ long and $3.5 \ m$ broad is carpeted with a carpet, leaving an uncovered margin $25 \ cm$ all around the room. If the breadth of the carpet is $80 \ cm$ , find its cost at $Rs. \ 60$ per meter.

Answer:

Dimensions of the Room: Length (l) $= 4.9 \ m$ , Breadth $(b) = 3.5 \ m$

Dimensions of the Carpet: Length (l) $= (4.9-0.5) = 4.4 \ m$ , Breadth $(b) = (3.5 - 0.5) = 3.0 \ m$

Area $= 4.4 \times 3 = 13.2 \ m^2$

Let the length of the carpet needed $= l$

Therefore $l \times 0.80 = 13.2$

$\Rightarrow l =$ $\frac{13.2}{0.80}$ $= 16.5 \ m$

Therefore cost $= 16.5 \times 0.60 = 9.9 \ Rs.$

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Question 34: The cost of carpeting a room at $Rs. \ 2.5$ Per square meter is $Rs. \ 450$ . The cost of whitewashing the walls at $50$ paisa per sq. meter is $Rs. \ 135$ . The room is $12 \ m$ wide. Find its height.

Answer:

$12 \times b \times 2.5 = 450$

$\Rightarrow b =$ $\frac{450}{12 \times 2.5}$ $= 15 \ m$

Cost of paining area of walls

$2 (l+b) h \times 0.5 = 135$

$\Rightarrow 2(12 + 15) h \times 0.5 = 135$

$\Rightarrow h =$ $\frac{135}{27}$ $= 5$

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Question 35: A room $5\ m$ long and $4 \ m$ wide is surrounded by a verandah. Find the width of the verandah if it occupies $22$ square meters.

Answer:

Let $x$ be the width of the varanda

Therefore $(5+2x)(4+2x) - 20 = 22$

$\Rightarrow 20 + 8x + 10x + 4x^2 - 20 = 22$

$\Rightarrow 4x^2 + 18x - 22 = 0$

$\Rightarrow (2x-2)(x +11) = 0$

$\Rightarrow x = 1$ or $x = -11$ (not possible)

Therefore the width of the verandah $= 1 \ m$

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Question 36: Square carpet is spread in the center of a room $55 \ dm$ square leaving a small margin of equal width all around. The total cost of carpeting at $25 \ paisa \ per \ square \ dm$ and decorating the margin at $15 \ paisa \ per \ square \ dm$ is $Rs. \ 703.75$ . Find the width of the margin.

Answer:

Dimension of square room $= 55 dm$

Let the width of the margin $= a$

Therefore area of carpet $= (55 - 2a)^2$

Area of margin $= 55^2 - (55 - 2a)^2$

$\Rightarrow 0.4a^2 - 22a + 52.5 = 0$

$\Rightarrow a^2 - 55a + 131.25 = 0$

$\Rightarrow (a-2.5)(a-52.5) = 0$

$\Rightarrow a = 2.5 \ or \ 52.5$ (not possible)

Hence the width of the margin $= 2.5 \ dm$

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Question 37: If the length and breadth of a room are increased by $1 \ m$ the area is increased By $21 \ m^2$ . If the length is increased by $1 \ m$ and breadth is decreased by $1\ m$ , the area is decreased by $5 \ m^2$ . Find the perimeter of the room.

Answer:

$(l+1)(b+1) - lb = 21$ … … … … … i)

$lb - (l+1)(b-1) = 5$ … … … … … ii)

From i) $lb + b + l +1 - lb = 21$

$\Rightarrow l+b = 20$ … … … … … iii)

Therefore Perimeter $= 2(l+b) = 2 \times 20 = 40 \ m$

from ii) $lb - (lb + b -l-1) = 5$

$\Rightarrow -b + l = 4$ … … … … … iv)

Solving (iii) and (iv) we get

$l = 12$ and $b = 8$ .

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Question 38: A rectangle has twice the area of a square. The length of the rectangle is $12 \ cm$ greater and the width is $8 \ cm$ greater than the side of the square. Find the perimeter of the square.

Answer:

Let the dimension of rectangle be $l$ and $b$ . And that of the square be $a$ .

Therefore $l = 12 + a$ and $b = 8 +a$

$(12+a)(8+a) = 2a^2$

$\Rightarrow96 + 8a + 12a + a^2 = 2a^2$

$\Rightarrow96 + 20a = a^2$

$\Rightarrow a^2 - 20a-96 = 0$

$\Rightarrow(a-24)(a+4) = 0$

$\Rightarrow a = 24 \ cm$

Therefore Perimeter of square $= 4 \times 24 = 96 \ cm$

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Question 39: If the perimeter of a rectangular plot is $58 \ m$ and the length of its diagonal is $26 \ m$ , find its area.

Answer:

Let the dimension of rectangle be $l$ and $b$ .

Therefore $2(l+b) = 68 \Rightarrow l+b = 34$ … … … … … i)

$\sqrt{l^2 + b^2} = 26$ … … … … … ii)

$(l+b)^2 = 1156$

$l^2 + b^2 +2 lb = 1156$

Therefore $l^2 + b^2 = 1156 - 2lb$

Substituting in (ii)

$\sqrt{1156-2lb} = 26$

$\Rightarrow 480 = 2lb \ or \ lb = 240$

Hence the area is $240 \ cm^2$

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Question 40: The length of a rectangular garden is $12 \ m$ more than its breadth. The numerical value of its area is equal to $4$ times the numerical value of its perimeter. Find the dimensions of its garden.

Answer:

Area $= (b+12)b$

Perimeter $= 2 (b+12+b) = 2(2b+12)$

Therefore $(b+12)b = 4 \times 2 (2b+12)$

$\Rightarrow b^2 + 12b = 16b + 96$

$\Rightarrow b^2 - 4b - 96 = 0$

$\Rightarrow (b-12)(b+8) = 0$

Therefore $b = 12$ or $b = -8$ (not possible)

Hence $b = 12$ . Therefore dimensions are $24 \ cm$ and $12 \ cm$

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Question 41: A wire when bent in the form of an equilateral triangle encloses an area $36\sqrt{6} \ cm^2$ of Find the area enclosed by the same wire when bent to form: (i) a square (ii) a rectangle whose length is $2 \ cm$ more than its width.